Question

The acceleration due to gravity on the earth’s surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is : ($h<<R$, where $R$ is the radius of the earth)

• A. $\frac{R^2{\omega }^2}{g}$
• B. $\frac{R^2{\omega }^2}{8g}$
• C. $\frac{R^2{\omega }^2}{4g}$
• D. $\frac{R^2{\omega }^2}{2g}$

Answer 1

The correct option is D

$\frac{R^2{\omega }^2}{2g}$

Step 1: Given data

Acceleration due to gravity on the earth’s surface at the poles $g$.

Angular velocity of the earth about the axis passing through the pole $\omega$.

Let the weight of an object at the equator $W_e$.

The weight of an object at the height $h$ above the poles is $W_h$.

It is given that,

$W_e=W_h$

Which implies,

$g_e=g_h$

Step 2: Formula used:

At equator,

Acceleration due to gravity will be-

$g_e=g-R{\omega }^2$

Where,$R$ is the radius of the earth

And at the height $h$, the acceleration due to gravity is $g_h=g(1-\frac{2h}{R})$

Step 3: Determine the height

As it is given that,

$g_e=g_h$.

So, on putting the values, we get-

$$\begin{gathered} g-R{\omega }^2=g\left(1-\frac{2h}{R}\right) \\ R{\omega }^2=g\left(1-1+\frac{2h}{R}\right) \\ R{\omega }^2=\frac{2gh}{R} \\ h=\frac{R^2{\omega }^2}{2g} \end{gathered}$$

Thus, when the weight is the same, the height will be $\frac{R^2{\omega }^2}{2g}$ .

Hence, option D is the correct answer.